Theory of Computing Systems

, Volume 58, Issue 2, pp 377–379

# Erratum to: “Top-down Tree Transducers with Regular Look-ahead”

Erratum
In Theorem 3.2 of my paper  it is shown that $$\textit {DB-FST}\subset \textit {DT}^{\mathit {R}}\textit {-FST}$$, where DB-FST is the class of deterministic bottom-up finite state tree transformations and DT R -FST is the class of deterministic top-down finite state tree transformations with regular look-ahead (and $$\subset$$ denotes proper inclusion). One may then ask for a characterization of DB-FST in terms of dt r -fst. In Theorem 3.2 of  it is wrongly stated that DB-FST = ODT R -FST, where ODT R -FST is the class of one-state dt r -fst. The correct statement is that
$$\textit{DB-FST}=\textit{FTA}\circ\textit{ODT}^{\mathit{R}}\textit{-FST},$$
where FTA is the class of tree transformations that are the identity on a recognizable (= regular) tree language. In other words, a tree transformation is a db-fst if and only if it is the restriction of a one-state dt r -fst to a recognizable tree language. Note that this implies that a total function from T Σ to T Δ is in DB-FST if and only if it is in ODT R -FST (as mentioned in Section 2 of ).

### Why the proof is wrong.

The proof of the inclusion $$\textit {DB-FST}\subseteq \textit {ODT}^{\mathit {R}}\textit {-FST}$$ of Theorem 3.2 is based on the fact that the construction in the proof of Lemma 2.10(3) preserves the number of states. But that is not true because in the latter proof it should be assumed that the initial state q d of T does not occur in the right-hand side of any rule of T. This can indeed be assumed without loss of generality, but possibly increases the number of states by one.

### Why the result is wrong.

Since $$\textit {FTA}\subset \textit {DB-FST}$$, it suffices to show that FTA is not included in ODT R -FST. Suppose that TODT R -FST is the identity on the recognizable tree language {σ(t)∣tT Σ}, where Σ={τ, a}, σ and τ have rank 1, and a has rank 0. Let T have the unique state q, and let t 0T Σ be a tree of size larger than the size of the rules of T. Consider the computation of T with input (and output) σ(t 0). The first rule applied in this computation must be of the form 〈q(σ(x))→s, D〉 such that t 0D(x) and q(x) occurs in s. Thus, there must be a successful computation of T starting with q(t 0). That contradicts the fact that t 0 is not in the domain of T. Hence $$\textit {ODT}^{\mathit {R}}\textit {-FST}\subset \textit {DB-FST}$$, where the inclusion is proper.

### Why the new result holds.

From the inclusion $$\textit {ODT}^{\mathit {R}}\textit {-FST}\subset \textit {DB-FST}$$ it follows that $$\textit {FTA}\circ \textit {ODT}^{\mathit {R}}\textit {-FST}\subseteq \textit {FTA}\circ \textit {DB-FST}\subseteq \textit {DB-FST}\circ \textit {DB-FST}\subseteq \textit {DB-FST}$$ where the last inclusion is Theorem 4.6(2) of .

It remains to prove that $$\textit {DB-FST}\subseteq \textit {FTA}\circ \textit {ODT}^{\mathit {R}}\textit {-FST}$$. We will say that a db-fst 〈Σ,Δ, Q, Q d , R〉 is full if Q d = Q, i.e., every state is final. We will denote the class of all full db-fst by FDB-FST, and the class of all full deterministic bottom-up finite state relabelings by FDBQREL. Note that FDB-FST is the class of total functions from T Σ to T Δ in DB-FST.

Since the domain of a db-fst is recognizable (by Corollary 3.12 of ), it should be clear that $$\textit {DB-FST}\subseteq \textit {FTA}\circ \textit {FDB-FST}$$. Consequently, it now suffices to prove that $$\textit {FDB-FST}\subseteq \textit {ODT}^{\mathit {R}}\textit {-FST}$$. In fact, the wrong proof of the inclusion $$\textit {DB-FST}\subseteq \textit {ODT}^{\mathit {R}}\textit {-FST}$$ in the proof of Theorem 3.2 is valid for full db-fst. It follows from the proof of Theorem 3.15(3) of  that $$\textit {FDB-FST}\subseteq \textit {FDBQREL}\circ \textit {HOM}$$. Since the identity is in ODT R -FST, this is included in ODT R -FSTFDBQRELHOM. Now the (wrong) construction in the proof of Lemma 2.10(3), discussed above, is actually correct for FDBQREL. Since that construction, and the one in the proof of Lemma 2.9, preserves the number of states, we obtain that $$\textit {ODT}^{\mathit {R}}\textit {-FST}\circ \textit {FDBQREL}\circ \textit {HOM}\subseteq \textit {ODT}^{\mathit {R}}\textit {-FST}$$. This shows that $$\textit {FDB-FST}\subseteq \textit {ODT}^{\mathit {R}}\textit {-FST}$$ and hence DB-FST = FTAODT R -FST.

### Remark.

Let O’DT R -FST be the class of dt r -fst with two states, such that the initial state does not occur in the right-hand side of any rule. It is not difficult to show that $$\textit {DB-FST}\subseteq \textit {O'DT}^{\mathit {R}}\textit {-FST}$$. In fact, the wrong proof of the inclusion $$\textit {DB-FST}\subseteq \textit {ODT}^{\mathit {R}}\textit {-FST}$$ in the proof of Theorem 3.2 is valid for O’DT R -FST, because the constructions in the proofs of Lemmas 2.10(3) and 2.9 preserve O’DT R -FST. It is also easy to see that the inclusion is proper, i.e., $$\textit {DB-FST}\subset \textit {O'DT}^{\mathit {R}}\textit {-FST}$$, as witnessed by the tree transformation {(t, σ(t))∣tT Σ}.

### Three other small corrections.

Six lines above Theorem 2.6 of  it is stated that $$\textit {ZT-FST}\subset \textit {ZT-FST}$$; that should of course be $$\textit {ZT-FST}\subset \textit {ZT}^{\mathit {R}}\textit {-FST}$$. On the 4th line of the proof of Lemma 2.10 of , the equation K = T L should be K = TL. After Theorem 2.11 of  it is stated that the inclusion signs in Theorem 2.6 may be replaced by equality signs; to see this, one should note that $$\textit {DBQREL}\subseteq \textit {LDT}^{\mathit {R}}\textit {-FST}$$ (by the proof of Lemma 2.10(3)).

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